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```				   in triangle ABC , D is the midpoint of AB . Pt.E is on AC such that AE=2EC . BE intersects CD at pt.F
Prove that BE=4EF
```

6 years ago

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```										Dear  ajinkya
Let A be (0,0),  B(x1,y1),   C(x2,y2)
then  By section formula
D=(x1/2,y1/2)
E=(2x2/3,2y2/3)
Let F(x3,y3)
let  BF:FE=k:1    and DF:FC=m:1
then by section formula on line BE and  CD we get
on BE........x3=(k*(2x2/3)+x1)/(k+1)   and y3= (k*(2y2/3)+y1)/(k+1)            ......(1)
On CD.......x3=(m*x2+x1/2)/(m+1)   and y3=(m*y2+y1/2)/(m+1)                         .........(2)
from (1) and (2)
(k*(2x2/3)+x1)/(k+1)  =(m*x2+x1/2)/(m+1)        and     (k*(2y2/3)+y1)/(k+1)  =(m*y2+y1/2)/(m+1)
Compare to  get coefficients of x1 and x2 or y1 and  y2
we get
2k/3(k+1)=m/(m+1)
and  1/(k+1)=1/2(m+1)
solve these two eq. to get k=3;
So BF:FE=3:1
==>BE:FE=4:1
==>BE=4EF
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best ajinkya !!!Regards,Askiitians ExpertsSahil AroraIIT Delhi
```
6 years ago

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